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Today's Topics:
1. Re: type problems with my little code (Gesh)
2. Re: About scope range: outside of the do-block (Gesh)
3. Re: type problems with my little code (Kim-Ee Yeoh)
4. Re: type problems with my little code (David McBride)
5. Re: graphical output? (John M. Dlugosz)
----------------------------------------------------------------------
Message: 1
Date: Sun, 20 Apr 2014 17:01:33 +0300
From: Gesh <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] type problems with my little code
Message-ID: <[email protected]>
Content-Type: text/plain; charset=UTF-8
On April 20, 2014 3:35:52 PM GMT+03:00, raffa f <[email protected]> wrote:
>hi everyone, i'm a complete beginner on programming and i've been
>solving
>the 99 haskell problems and i've come into a situation. just like all
>my
>problems with haskell, it's about types. i wrote a little thing that
>takes
>a list and puts all the equal characters together, like [1,1,3,2,3]
>will be
>[ [1,1], [3,3], [2] ]. here it is:
>
>pack xs = pack' xs
>
> where
>
> pack' :: [a] -> [[a]]
>
> pack' [] = []
>
> pack' (x:y) = extra x xs ++ pack' y
>
>
>
>extra x xs = [sum' (map (remover x) (map ((==) x) xs))]
>
>
>
>remover :: a -> Bool -> [a]
>
>remover y x = if x then [y] else []
>
>
>
>sum' :: [[a]] -> [a]
>
>sum' [] = []
>
>sum' (x:xs) = x ++ sum' xs
>
>
>i know, i know, this code is probably terrible and i'm sure there are
>more
>clever ways to do this...but i wanna understand what's wrong. the
>"extra"
>function works perfectly, it takes a variable and then looks at how
>many
>times it's presented on a list and outputs a list with a list of that
>variable the amount of times that it was presented on the original
>list.
>
>however, pack does not work quite right. first of all, it'll use extra
>on
>repeated characters, creating repeated lists, but that doesn't matter
>because i have a function that fixes that issue and i'll use it after i
>figure out what's wrong with pack. now, the real problem is with types.
>here's what haskell says when i try to load it:
>
>noventa.hs:77:32:
> Couldn't match type `a' with `a1'
> `a' is a rigid type variable bound by
> the inferred type of pack :: [a] -> [[a]] at noventa.hs:73:1
> `a1' is a rigid type variable bound by
> the type signature for pack' :: [a1] -> [[a1]] at
>noventa.hs:75:19
> Expected type: [a1]
> Actual type: [a]
> In the second argument of `extra', namely `xs'
> In the first argument of `(++)', namely `extra x xs'
> In the expression: extra x xs ++ pack' y
>Failed, modules loaded: none.
>
>
>i don't understand this at all! if i replace [a] with String and [[a]]
>with
>[String], it works! but i want pack to work for lists of numbers too...
>types are so confusing. can anyone help me?
>
>
>------------------------------------------------------------------------
>
>_______________________________________________
>Beginners mailing list
>[email protected]
>http://www.haskell.org/mailman/listinfo/beginners
I'm a bit fuzzy on this part of GHC's type checker, so if someone could correct
me, that'd be awesome.
If I understand correctly, what's happening here is that the type signature you
gave pack' implicitly told GHC that pack' can work with any list, independently
of the type of xs.
However, this is incorrect, as the call to extra forces the type of the input
to pack' to be a list with the same type of elements as those of xs. Thus, GHC
complains that you told it pack' accepts more stuff than it actually does.
This can be fixed in one of a few ways - this list might not be exhaustive,
though.
- Keep the type of pack' as it currently is, but fix its implementation to
actually have that type. Since our problem was caused by passing xs to extra
restricting the type, you'd need to replace xs in that expression by something
else, e.g. x:y
- Fix the type of pack' to what you actually meant. This can be done either by
removing the signature, allowing GHC to fill in the correct type, or by turning
on ScopedTypeVariables, writing pack's signature and making sure you use the
same type variable for the elements of the parameters of pack and pack'.
Besides all of the above, there are a few more comments I'd like to make on
your code.
First, note that sum' = concat
Next, extra x xs = [filter (==x) xs]
Then, noting that [x] ++ xs = x:xs, you obtain that
pack' (x:xs) = filter (==x) xs : pack' xs
In addition, the type signature you gave pack' lacks an Eq a constraint, which
is needed as you are testing the elements of the parameter for equality.
Finally, pack doesn't do what you want it to do, as it will replace every
element of its parameter by a list containing as many occurrences of that
element as there are occurrences of that element in the parameter. Can you
think of a way to solve this problem?
Good luck learning! I hope you'll find this journey interesting and beautiful.
Gesh
------------------------------
Message: 2
Date: Sun, 20 Apr 2014 17:06:30 +0300
From: Gesh <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] About scope range: outside of the
do-block
Message-ID: <[email protected]>
Content-Type: text/plain; charset=UTF-8
On April 20, 2014 4:38:05 PM GMT+03:00, "S. H. Aegis" <[email protected]> wrote:
>Oh, I'm sorry.
>I delete some of commented lines.
>Sorry for confusing.
>
>Error occurred in
>sortedNamesList = sort $ nameList names
>
>Thank you.
>
>
>2014-04-20 22:17 GMT+09:00 Norbert Melzer <[email protected]>:
>
>>
>> Am 20.04.2014 14:58 schrieb "S. H. Aegis" <[email protected]>:
>>
>> > namesScores.hs:31:35: Not in scope: `names'
>> > Failed, modules loaded: none.
>>
>> The code you posted has only 21 lines while your error message
>denotes an
>> error in line 31, so there is something missing.
>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://www.haskell.org/mailman/listinfo/beginners
>>
>>
In general, variables defined locally in one scope are not floated to external
scopes.
Thus, your question is analogous to asking:
Given the following code:
foo = ...
where bar = ...
baz = ...
Can I call bar from baz? The answer being no, you can't, unless you pass bar as
a parameter to baz (or use a Reader monad, which is equivalent).
Hoping to help,
Gesh
------------------------------
Message: 3
Date: Sun, 20 Apr 2014 21:24:44 +0700
From: Kim-Ee Yeoh <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] type problems with my little code
Message-ID:
<capy+zdq82qce-he+w5gwuvyocjevd+xmcyjy9ugg_lpjvqq...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
First of all, Raffa, welcome to the list. I hope your stay is friendly and
that you'll derive benefit from it over the long-term.
This line that you wrote
pack' (x:y) = extra x xs ++ pack' y
suggests that you might want to review the different between list-consing,
which is the (:) function, and list-append, which is (++).
Especially for Haskell, you'll need to pay close attention to how the type
signatures differ.
These concepts are very, very old and easily google-able.
A solid foundation in these fundamentals will make tackling the rest of the
99 problems so much easier.
-- Kim-Ee
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Message: 4
Date: Sun, 20 Apr 2014 10:56:50 -0400
From: David McBride <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] type problems with my little code
Message-ID:
<CAN+Tr41aecsaE1cpw7ML=jldpe9wdetrde-sfvvsz+sgxjh...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
You are running into a quirk in the type system. The haskell standard says
that sub expressions in where clauses that have type annotations with
variables in them (pack'), the variables are not the same as the variables
in the toplevel expression (pack). When you wrote this code, haskell
inferred a type of pack :: (Eq a1) => [a1] -> [[a1]], and then pack' is
correctly the type pack' :: [a] -> [[a]], but that a1 and that a are
different types as far as ghc is concerned. But based on how they are
used, ghc says no, they should be the same type, ERROR FIX IT. Even if you
specify a type for pack that uses "a", the problem still persists because
they are still considered different a's.
There are two ways to fix this. First one is just remove the type info
from pack'. Just remove that line and ghc will infer all the types
correctly and you can get on with your day.
The other way is to use the ScopedTypeVariables extension, which everyone
generally loves and uses when they need it. It has a niggle where it
requires a forall for each type variable in the overall type, but from that
point on it works the way you would expect.
{-# LANGUAGE ScopedTypeVariables #-}
pack :: forall a. (Eq a) => [a] -> [[a]]
pack xs = pack' xs
where
pack' :: [a] -> [[a]]
pack' [] = []
pack' (x:y) = extra x xs ++ pack' y
On Sun, Apr 20, 2014 at 8:35 AM, raffa f <[email protected]> wrote:
> hi everyone, i'm a complete beginner on programming and i've been solving
> the 99 haskell problems and i've come into a situation. just like all my
> problems with haskell, it's about types. i wrote a little thing that takes
> a list and puts all the equal characters together, like [1,1,3,2,3] will be
> [ [1,1], [3,3], [2] ]. here it is:
>
> pack xs = pack' xs
>
> where
>
> pack' :: [a] -> [[a]]
>
> pack' [] = []
>
> pack' (x:y) = extra x xs ++ pack' y
>
>
>
> extra x xs = [sum' (map (remover x) (map ((==) x) xs))]
>
>
>
> remover :: a -> Bool -> [a]
>
> remover y x = if x then [y] else []
>
>
>
> sum' :: [[a]] -> [a]
>
> sum' [] = []
>
> sum' (x:xs) = x ++ sum' xs
>
>
> i know, i know, this code is probably terrible and i'm sure there are more
> clever ways to do this...but i wanna understand what's wrong. the "extra"
> function works perfectly, it takes a variable and then looks at how many
> times it's presented on a list and outputs a list with a list of that
> variable the amount of times that it was presented on the original list.
>
> however, pack does not work quite right. first of all, it'll use extra on
> repeated characters, creating repeated lists, but that doesn't matter
> because i have a function that fixes that issue and i'll use it after i
> figure out what's wrong with pack. now, the real problem is with types.
> here's what haskell says when i try to load it:
>
> noventa.hs:77:32:
> Couldn't match type `a' with `a1'
> `a' is a rigid type variable bound by
> the inferred type of pack :: [a] -> [[a]] at noventa.hs:73:1
> `a1' is a rigid type variable bound by
> the type signature for pack' :: [a1] -> [[a1]] at
> noventa.hs:75:19
> Expected type: [a1]
> Actual type: [a]
> In the second argument of `extra', namely `xs'
> In the first argument of `(++)', namely `extra x xs'
> In the expression: extra x xs ++ pack' y
> Failed, modules loaded: none.
>
>
> i don't understand this at all! if i replace [a] with String and [[a]]
> with [String], it works! but i want pack to work for lists of numbers
> too... types are so confusing. can anyone help me?
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
>
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------------------------------
Message: 5
Date: Sun, 20 Apr 2014 23:11:18 -0500
From: "John M. Dlugosz" <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] graphical output?
Message-ID: <[email protected]>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
On 4/20/2014 1:31 AM, Bill Noble wrote:
> (note for some reason I had to install llvm first).
>
I had already tried that, and ran into the same problem.
------------------------------
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